Problem: Simplify the following expression: $y = \dfrac{7x^2+18x+8}{7x + 4}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(8)} &=& 56 \\ {a} + {b} &=& &=& {18} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $56$ and add them together. The factors that add up to ${18}$ will be your ${a}$ and ${b}$ When ${a}$ is ${4}$ and ${b}$ is ${14}$ $ \begin{eqnarray} {ab} &=& ({4})({14}) &=& 56 \\ {a} + {b} &=& {4} + {14} &=& 18 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 +{4}x) + ({14}x +{8}) $ Factor out the common factors: $ x(7x + 4) + 2(7x + 4)$ Now factor out $(7x + 4)$ $ (7x + 4)(x + 2)$ The original expression can therefore be written: $ \dfrac{(7x + 4)(x + 2)}{7x + 4}$ We are dividing by $7x + 4$ , so $7x + 4 \neq 0$ Therefore, $x \neq -\frac{4}{7}$ This leaves us with $x + 2; x \neq -\frac{4}{7}$.